Assume X Has the Following Probability Distribution $$\begin{array} { l l l l l }
X & 1 & 2 & 3 & 4 \
P ( X ) & .1 & .5 & .2 & .2
\end{array}$$

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Assume X Has the Following Probability Distribution $$\begin{array} { l l l l l } 
X & 1 & 2 & 3 & 4 \
P ( X ) & .1 & .5 & .2 & .2
\end{array}$$

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The Answer of Assume X Has the Following Probability Distribution $$\begin{array} { l l l l l } 
X & 1 & 2 & 3 & 4 \
P ( X ) & .1 & .5 & .2 & .2
\end{array}$$

Assume X Has the Following Probability Distribution $\begin{array} { l l l l l } X & 1 & 2 & 3 & 4 \\P ( X ) & .1 & .5 & .2 & .2\end{array}$

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Assume X Has the Following Probability Distribution X1234P(X).1.5.2.2\begin{array} { l l l l l } X & 1 & 2 & 3 & 4 \\P ( X ) & .1 & .5 & .2 & .2\end{array}XP(X)​1.1​2.5​3.2​4.2​

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Assume X Has the Following Probability Distribution $\begin{array} { l l l l l } X & 1 & 2 & 3 & 4 \\P ( X ) & .1 & .5 & .2 & .2\end{array}$

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Verified
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